/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 95 One end of a long glass rod is g... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

One end of a long glass rod is ground to a convex hemispherical shape. This glass an index of refraction of \(1.55 .\) When a small leaf is placed 20.0 \(\mathrm{cm}\) in front of the center of the hemisphere along the optic axis, an image is formed inside the glass 9.12 \(\mathrm{cm}\) from the spherical surface. Where would the image be formed if the glass were now immersed in water (refractive index 1.33\()\) but nothing else were changed?

Short Answer

Expert verified
The image is formed farther away from the object compared to the initial configuration.

Step by step solution

01

Identify and Write Down Known Values

We have the index of refraction of the glass, \( n_1 = 1.55 \), and the initial position of the object \( s_1 = -20.0 \text{ cm} \) (negative because the object is outside the surface towards the light source). The position of the image inside the glass initially is \( s_1' = 9.12 \text{ cm} \). For water, the index of refraction \( n_2 = 1.33 \).
02

Calculate the Radius of Curvature

Using the refraction formula for a spherical surface: \[ \frac{n_1}{s_1} + \frac{n_2}{s_1'} = \frac{n_2 - n_1}{R} \]Substitute \( n_1=1 \), \( n_2=1.55 \), and solve for \( R \):\[ \frac{1}{-20} + \frac{1.55}{9.12} = \frac{1.55 - 1}{R} \]Calculate \( R \).
03

Image Formation in Water

Now, the glass is immersed in water. We apply the spherical refractor formula again using \( n_2 = 1.33 \) and \( n_1 = 1.55 \):\[ \frac{1.33}{s_2} + \frac{1.55}{s_2'} = \frac{1.55 - 1.33}{R} \]Solve for \( s_2' \), the new image distance in the water.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Spherical surfaces
A spherical surface can be described as part of a sphere that is used in lenses and other optical instruments. These surfaces are crucial in focusing or spreading light rays.
Spherical surfaces have two main types: convex and concave. A convex spherical surface curves outward, like the outside of a sphere. This type is common in lenses that converge light.
Conversely, a concave surface curves inward, like the inside of a spoon, and is typically used in lenses that diverge light.
Spherical surfaces play an important role in optics, influencing how light is bent or refracted when it hits the surface. This bending of light ultimately forms images at certain positions relative to the surface.
Index of refraction
The index of refraction defines how much slower light travels in a medium compared to a vacuum. It's a measure used to indicate the speed reduction of light as it passes through different materials.
Every material has its unique index of refraction. For example, air has an index close to 1, meaning light travels almost as fast as it does in a vacuum. Glass, however, has a higher index, such as 1.55, indicating that light travels slower in glass than in air.
This concept is crucial when considering how light bends at the interface of two materials. The greater the difference in indices between two media, the more the light will bend when transitioning from one medium to the other. This principle is essential for understanding image formation through lenses or other curved optical elements.
Radius of curvature
The radius of curvature is the distance from the center of curvature to the surface itself. In the context of spherical surfaces, this measurement describes the "size" of the sphere from which the surface is derived.
If you imagine a complete sphere from which a lens or surface is made, the radius of curvature is the radius of that larger sphere.
Understanding the radius of curvature is vital in optics because it affects how lenses bend light. A surface with a smaller radius of curvature is more curved, which can lead to stronger lens effects, like more focusing power. Conversely, a larger radius results in a flatter surface, reducing the bending of light and affecting the focal length of lenses or mirrors.
Optical axis
The optical axis of a lens or optical system is an imaginary line that defines the path along which light travels without being refracted. This line extends from the object through the lens to the point where images form.
In symmetric optical devices, like a spherical lens, the optical axis passes through the center of curvature and the vertex of the lens. It is a fundamental component to consider when aligning optical elements.
The significance of the optical axis increases when calculating image distances, as it helps determine the shortest path light travels through the lens. Proper alignment along the optical axis ensures more accurate image formation, reducing errors or distortions in optical systems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A thick-walled wine goblet sitting on a table can be considered to be a hollow glass sphere with an outer radius of 4.00 cm and an inner radius of 3.40 \(\mathrm{cm}\) . The index of refraction of the goblet glass is \(1.50 .(a)\) A beam of parallel light rays enters the side of the empty goblet along a horizontal radius. Where, if anywhere, will an image be formed? ( \(b\) ) The goblet is filled with white wine \((n=1.37)\) . Where is the image formed?

When an object is placed at the proper distance to the left of a converging lens, the image is focused on a screen 30.0 \(\mathrm{cm}\) to the right of the lens. A diverging lens is now placed 15.0 \(\mathrm{cm}\) to the right of the converging lens, and it is found that the screen must be moved 19.2 \(\mathrm{cm}\) farther to the right to obtain a sharp image. What is the focal length of the diverging lens?

A solid glass hemisphere of radius 12.0 \(\mathrm{cm}\) and index of refraction \(n=1.50\) is placed with its flat face downward on a table. A parallel beam of light with a circular cross section 3.80 \(\mathrm{mm}\) in diameter travels straight down and enters the hemisphere at the center of its curved surface. (a) What is the diameter of the circle of light formed on the table? (b) How does your result depend on the radius of the hemisphere?

If you run away from a plane mirror at 2.40 \(\mathrm{m} / \mathrm{s}\) , at what speed does your image move away from you?

Dental Mirror. A dentist uses a curved mirror to view teeth on the upper side of the mouth. Suppose she wants an enect image with a magnification of 2.00 when the mirror is 1.25 \(\mathrm{cm}\) from a tooth. (Treat this problem as though the object and image lie along a straight line.) (a) What kind of mirror (concave or convex) is needed? Use a ray diagram to decide, without performing any calculations. (b) What must be the focal length and radius of curvature of this mirror?(c) Draw a principal-ray diagram to check your answer in part (b).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.